Let $T,U:V\to W$ be linear transformations. Prove that if $W$ is finite-dimensional, then $\text{rank}(T+U)\leq\text{rank}(T) + \text{rank}(U)$.

Question

Let $T,U:V\to W$ be linear transformations.

(a) Prove that $R(T+U)\subseteq R(T) + R(U)$.

(b) Prove that if $W$ is finite-dimensional, then $\text{rank}(T+U)\leq\text{rank}(T) + \text{rank}(U)$.

MY ATTEMPT

(a) If $w\in R(T+U)$, there exists $v\in V$ such that $w = (T+U)v = T(v) + U(v)$.

But $T(v)\in R(T)$ and $U(v)\in R(U)$.

Thus $w\in R(T) + R(U)$, from whence we conclude that $R(T+U)\subseteq R(T)+R(U)$.

(b) Since $R(T + U)$ is a linear subspace of $R(T)+R(U)$, we have that
\begin{align*}
\dim R(T+U) & \leq \dim(R(T) + R(U)) = \dim R(T) + \dim R(U) – \dim(R(T)\cap R(U))\\\\
& \leq \dim R(T) + \dim R(U)
\end{align*}

Could someone verify if I am not doing any conceptual mistake?

Answer 1

$a)$ looks good.

$b)$ follows from $a)$, since $\operatorname{rank}T=\operatorname{dim}R(T)$

(You correctly used the formula for the dimension of the sum of two subspaces.)